# The beam of maximal cross sectional area is square.

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

### Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

#### Solutions

Chapter 4, Problem 63RE

(a)

To determine

## To find: The beam of maximal cross sectional area is square.

Expert Solution

The beam of maximal cross sectional area is square.

A=x4r2x2

### Explanation of Solution

Given:

r=10inches

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

A=x4r2x2

(b)

To determine

### To find: The dimensions of the planks that will have maximal cross-sectional area.

Expert Solution

The dimensions of the planks that will have maximal cross-sectional area is x=2r .

### Explanation of Solution

Given:

r=10inches

Concept used:

Pythagoras theorem :- In triangle, the sum of the squares of the lengths of the triangles legs is the same as the square of the length of the triangle hypotenuse.

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

A=x4r2x2...............................(2)

Differentiating with respect to x

Using the zero product property of the equation

0=4r2x22x24r2x20=4r2x22x20=4r2x2x=±2rx=2r

(c)

To determine

### To find: The dimensions of the strongest beam that cut from the cylindrical log.

Expert Solution

The dimensions of the strongest beam that cut from the cylindrical log is x4r2x2=y

### Explanation of Solution

Given:

r=10inches

Concept used:

If the quantities x and y are related by an equation

yαx

For some constant k0

That is y varies inversely proportional as x or y is directly proportional to x

The constant k is called constant proportionality

Calculation:

Let x be depth of the rectangle and y be width of the rectangle.

The value of depth and width are equal then area is square.

Form the given data the diagonal of the rectangle length 2r .

By using Pythagoras theorem

x2+y2=(2r)2y2=(2r)2x2y=4r2x2

The area of the rectangule A=xy................................(1)

Now after substituting 4r2x2 for y in equation (1)

A=x4r2x2

If the quantities x and y are related by an equation

yαxy=kxx4r2x2=y

For some constant k0

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